Fourth Edition CHAPTER 8 MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University Principle Stresses Under a Given Loading © 2006 The McGraw-Hill Companies, Inc. All rights reserved. Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Principle Stresses Under a Given Loading Introduction Principle Stresses in a Beam Sample Problem 8.1 Sample Problem 8.2 Design of a Transmission Shaft Sample Problem 8.3 Stresses Under Combined Loadings Sample Problem 8.5 © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8-2 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Introduction • In Chapter 1 and 2, you learned how to determine the normal stress due to centric loads. • In Chapter 3, you analyzed the distribution of shearing stresses in a circular member due to a twisting couple. • In Chapter 4, you determined the normal stresses caused by bending couples. • In Chapters 5 and 6, you evaluated the shearing stresses due to transverse loads. • In Chapter 7, you learned how the components of stress are transformed by a rotation of the coordinate axes and how to determine the principal planes, principal stresses, and maximum shearing stress at a point. • In Chapter 8, you will learn how to determine the stress in a structural member or machine element due to a combination of loads and how to find the corresponding principal stresses and maximum shearing stress. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8-3 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Principle Stresses in a Beam • Prismatic beam subjected to transverse loading My Mc σm = I I VQ VQ τ xy = − τm = It It σx = − • Principal stresses determined from methods of Chapter 7 • Can the maximum normal stress within the cross-section be larger than σm = © 2006 The McGraw-Hill Companies, Inc. All rights reserved. Mc I 8-4 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Principle Stresses in a Beam © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8-5 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Principle Stresses in a Beam • Cross-section shape results in large values of τxy near the surface where σx is also large. • σmax may be greater than σm © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8-6 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.1 SOLUTION: • Determine shear and bending moment in Section A-A’ • Calculate the normal stress at top surface and at flange-web junction. A 160-kN force is applied at the end of a W200x52 rolled-steel beam. • Evaluate the shear stress at flangeweb junction. Neglecting the effects of fillets and of stress concentrations, determine whether the normal stresses satisfy a design specification that they be equal to or less than 150 MPa at section A-A’. • Calculate the principal stress at flange-web junction © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8-7 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.1 SOLUTION: • Determine shear and bending moment in Section A-A’ M A = (160 kN )( 0.375 m ) = 60 kN - m V A = 160 kN • Calculate the normal stress at top surface and at flange-web junction. MA 60 kN ⋅ m = S 512 × 10−6 m3 = 117.2 MPa y 90.4 mm σb = σ a b = (117.2 MPa ) c 103 mm = 102.9 MPa σa = © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8-8 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.1 • Evaluate shear stress at flange-web junction. Q = ( 204 × 12.6 ) 96.7 = 248.6 × 103 mm3 = 248.6 × 10−6 m3 ( ) V AQ (160 kN ) 248.6 × 10−6 m3 τb = = It 52.7 × 10− 6 m 4 ( 0.0079 m ) = 95.5 MPa ( ) • Calculate the principal stress at flange-web junction σ max = 12 σ b + ( 12 σ b )2 + τ b2 2 102.9 102.9 2 = + + ( 95.5) 2 2 = 159.9 MPa ( > 150 MPa ) Design specification is not satisfied. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8-9 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.2 SOLUTION: • Determine reactions at A and D. • Determine maximum shear and bending moment from shear and bending moment diagrams. The overhanging beam supports a uniformly distributed load and a concentrated load. Knowing that for the grade of steel to used σall = 24 ksi and τall = 14.5 ksi, select the wideflange beam which should be used. • Calculate required section modulus and select appropriate beam section. • Find maximum normal stress. • Find maximum shearing stress. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 10 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.2 SOLUTION: • Determine reactions at A and D. ∑ M A = 0 ⇒ RD = 59 kips ∑ M D = 0 ⇒ R A = 41kips • Determine maximum shear and bending moment from shear and bending moment diagrams. M max = 239.4 kip ⋅ in V max = 43 kips with V = 12.2 kips • Calculate required section modulus and select appropriate beam section. M max 24 kip ⋅ in S min = = = 119.7 in 3 σ all 24 ksi select W21× 62 beam section © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 11 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.2 • Find maximum shearing stress. Assuming uniform shearing stress in web, V 43 kips τ max = max = = 5.12 ksi < 14.5 ksi Aweb 8.40 in 2 • Find maximum normal stress. σa = M max 60 kip ⋅ in = 2873 = 22.6 ksi 3 S 127in y 9.88 σb = σ a b = ( 22.6 ksi ) = 21.3 ksi c 10.5 τb = V 12.2 kips = = 1.45 ksii Aweb 8.40 in 2 2 21.3 ksi 21.3 ksi 2 σ max = + + (1.45 ksi ) 2 2 = 21.4 ksi < 24 ksi © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 12 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Design of a Transmission Shaft • If power is transferred to and from the shaft by gears or sprocket wheels, the shaft is subjected to transverse loading as well as shear loading. • Normal stresses due to transverse loads may be large and should be included in determination of maximum shearing stress. • Shearing stresses due to transverse loads are usually small and contribution to maximum shear stress may be neglected. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 13 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Design of a Transmission Shaft • At any section, Mc I Tc τm = J σm = where M 2 = M y2 + M z2 • Maximum shearing stress, 2 2 σ Mc Tc τ max = m + (τ m ) 2 = + 2 2I J 2 for a circular or annular cross - section, 2 I = J τ max = c M2 +T2 J • Shaft section requirement, M 2 + T 2 max J = τ all c min © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 14 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.3 SOLUTION: • Determine the gear torques and corresponding tangential forces. • Find reactions at A and B. • Identify critical shaft section from torque and bending moment diagrams. Solid shaft rotates at 480 rpm and transmits 30 kW from the motor to gears G and H; 20 kW is taken off at gear G and 10 kW at gear H. Knowing that σall = 50 MPa, determine the smallest permissible diameter for the shaft. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. • Calculate minimum allowable shaft diameter. 8 - 15 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.3 SOLUTION: • Determine the gear torques and corresponding tangential forces. P 30 kW = = 597 N ⋅ m 2πf 2π ( 8 Hz ) T 597 N ⋅ m FE = E = = 3.73 kN rE 0.16 m TE = 20 kW = 398 N ⋅ m 2π ( 8 Hz ) 10 kW TD = = 199 N ⋅ m 2π ( 8 Hz ) TC = FC = 6.63 kN FD = 2.49 kN • Find reactions at A and B. Ay = 0.932 kN Az = 6.22 kN B y = 2.80 kN Bz = 2.90 kN © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 16 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.3 • Identify critical shaft section from torque and bending moment diagrams. ( M y2 + M z2 + T 2 ) max = 11602 + 3732 + 597 2 = 1357 N ⋅ m © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 17 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.3 • Calculate minimum allowable shaft diameter. M y2 + M z2 + T 2 J = c τ all = 1357 N ⋅ m = 27.14 × 10− 6 m3 50 MPa For a solid circular shaft, J π 3 = c = 27.14 × 10− 6 m3 c 2 c = 0.02585 m = 25.85 m d = 2c = 51.7 mm © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 18 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Stresses Under Combined Loadings • Wish to determine stresses in slender structural members subjected to arbitrary loadings. • Pass section through points of interest. Determine force-couple system at centroid of section required to maintain equilibrium. • System of internal forces consist of three force components and three couple vectors. • Determine stress distribution by applying the superposition principle. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 19 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Stresses Under Combined Loadings • Axial force and in-plane couple vectors contribute to normal stress distribution in the section. • Shear force components and twisting couple contribute to shearing stress distribution in the section. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 20 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Stresses Under Combined Loadings • Normal and shearing stresses are used to determine principal stresses, maximum shearing stress and orientation of principal planes. • Analysis is valid only to extent that conditions of applicability of superposition principle and Saint-Venant’s principle are met. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 21 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.5 SOLUTION: • Determine internal forces in Section EFG. • Evaluate normal stress at H. • Evaluate shearing stress at H. Three forces are applied to a short steel post as shown. Determine the principle stresses, principal planes and maximum shearing stress at point H. • Calculate principal stresses and maximum shearing stress. Determine principal planes. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 22 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.5 SOLUTION: • Determine internal forces in Section EFG. Vx = −30 kN P = 50 kN Vz = −75 kN M x = ( 50 kN )( 0.130 m ) − ( 75 kN )( 0.200 m ) = −8.5 kN ⋅ m M y = 0 M z = ( 30 kN )( 0.100 m ) = 3 kN ⋅ m Note: Section properties, A = ( 0.040 m )( 0.140 m ) = 5.6 × 10−3 m 2 1 ( 0.040 m )( 0.140 m ) 3 = 9.15 × 10− 6 m 4 I x = 12 1 ( 0.140 m )( 0.040 m ) 3 = 0.747 × 10 − 6 m 4 I z = 12 © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 23 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.5 • Evaluate normal stress at H. σy =+ = P Mz a Mx b + − A Iz Ix ( 3 kN ⋅ m )( 0.020 m ) + 5.6 × 10-3 m 2 0.747 × 10− 6 m 4 50 kN − ( 8.5 kN ⋅ m )( 0.025 m ) 9.15 × 10−6 m 4 = ( 8.93 + 80.3 − 23.2 ) MPa = 66.0 MPa • Evaluate shearing stress at H. Q = A1 y1 = [ ( 0.040 m )( 0.045 m ) ] ( 0.0475 m ) = 85.5 × 10−6 m3 ( ) ( Vz Q 75 kN ) 85.5 × 10− 6 m3 τ yz = = I xt 9.15 × 10 −6 m 4 ( 0.040 m ) ( ) = 17.52 MPa © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 24 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.5 • Calculate principal stresses and maximum shearing stress. Determine principal planes. τ max = R = 33.02 + 17.522 = 37.4 MPa σ max = OC + R = 33.0 + 37.4 = 70.4 MPa σ min = OC − R = 33.0 − 37.4 = −7.4 MPa tan 2θ p = CY 17.52 = 2θ p = 27.96° CD 33.0 θ p = 13.98° τ max = 37.4 MPa σ max = 70.4 MPa σ min = −7.4 MPa θ p = 13.98° © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 8 - 25